Solutions to Homework 6 Mathematics 503
Foundations of Mathematics Spring 2014
§3.4: 1. If m is any integer, then m(m+ 1) =m2+m is the product of m and its successor.
That it to say, m2+m is the product of two consecutive integers. The results of the preview activities verify that this product is even by examining the two possibilities: m is odd, or m is even. In both cases, the product of m with m+ 1 is divisible by 2. 2. We prove the following proposition.
Proposition. Suppose that u is an odd integer; then the equation x2+x−u= 0 has
no integer solutions.
Proof. Suppose for contradiction that there exists a integer solution mof the equation x2+x−u= 0,
where uis an odd integer. We will examine the two cases: m is odd and m is even. If m is odd, then there exists an integer k such that
m= 2k+ 1. Since m is a root of our equation, it must be that
(2k+ 1)2+ (2k+ 1)−u= 0. This implies that
4k2+ 4k+ 1 + 2k+ 1−u= 0.
This is clearly not possible, for 4k2+ 10k+ 2 is even. Now suppose that m is even. If so, then there exists an integer k such thatm = 2k; then
4k2+ 2k−u= 0. This is also impossible, for 4k2+ 2k is even.
3. We prove the following.
Proposition. If n is an odd integer, the n = 4k+ 1 for some integer k or n= 4k+ 3 for some integer k.
Proof. Suppose that n is odd; then there is an integer j such that n = 2j + 1. There are two possibilities for j. Eitherj is even, or j is odd.
Ifj is even, then there is an integerk such thatj = 2k. Thus n= 4k+ 1.
On the other hand, ifj is odd, there is an integer k such thatj = 2k+ 1. This means that
n= 2(2k+ 1) + 1. This is equivalent to
n= 4k+ 3.
6. (a) We prove the following.
Proposition. If m and n are consecutive integers, then 4 divides m2+n2−1.
Proof. Suppose that m and n are consecutive integers; we can assume that n = m+ 1 in this situation. Then
m2+n2−1 = m2+ (m+ 1)2−1 = m2+m2+ 2m+ 1−1 = 2m2+ 2m= 2m(m+ 1). Now there are two cases: eitherm is even, or m is odd. If m is even, then 4 will divide 2m, and therefore 2m(m+ 1). Ifmis odd, thenm+ 1 is even. 2 will divide 2m, and 2 will also divide m+ 1. It follows that 4 will divide 2m(m+ 1).
On the other hand, the converse is untrue. Let m= 1 and let n= 4. (b) We prove the following.
Proposition. For all integers m, n, if m and n are both even or if m and n are both odd, then 4 divides m2−n2.
Proof. First suppose that bothm and n are even. If so, then there exist integers j and k such that m= 2j and n = 2j. So
m2 −n2 = 4k2−4j2 = 4(k2−j2). This implies that 4 dividesm2−n2.
Now suppose thatm and n are both odd. If so, then there exist integersk and j such thatm = 2k+ 1 andn = 2j+ 1. So
m2−n2 = 4k2+ 4k+ 1−4j2−4j−1 = 4(k2+k−j2−j) This means that 4 dividesm2−n2 in this case also.
Proposition. For all integers m, n, if 4 divides m2−n2, then either both m and n are even, or m and n are odd.
Proof. Consider the contrapositive. Suppose that m is odd and n is even. If so, then there are two integers j and k such thatm = 2j+ 1 and n= 2k. So
m2−n2 = (2j+ 1)2−4k2 = 4j2+ 4j + 1−4k2. This is an odd number, so is not divisible by 4.
Suppose thatm is even and n is odd. If so, then there are two integers j and k such thatm = 2j and n = 2k+ 1. So
m2−n2 = 4j2−4k2−4k−1. This number is also odd, so it is not divisible by 4. 7. We prove
Proposition. If n is an odd integer, then 8|n2−1.
Proof. If n is odd, then there is an integer k such that n = 2k+ 1. Then n2 −1 = (2k+ 1)2 −1 = 4k2 + 4k = 4(k2 +k). By a result proved in class (and appearing
elsewhere in this homework assignment), k2 +k must be even, i.e., k2 +k = 2j for
some integer j. Thus n2−1 = 4(k2+k) = 4(2j) = 8j, so 8|n2−1.
10. (b) We prove the following.
Proposition. For all real numbers x and y, |xy|=|x||y|.
Proof. There are several cases to consider. Clearly, if either x or y is equal to 0, then |xy| = |x||y|. The equation is clearly true if x > 0 and y > 0, for |xy| = xy = |x||y| in this case. Suppose that x > 0 and y < 0. Then, |xy|= −xy. On ther other hand, |x|=x and |y|=−y, so|x||y|=−xy as well. Similarly, the equation is true if x <0 and y > 0. Finally, consider what happens when x < 0 and y < 0. In this case, the productxy >0, so|xy|=xy. On the other hand,|x|=−xand|y|=−y. This implies that |x||y|=xy as well.
13. (a) This proof is incorrect, but the statement is true. It correctly deduces that an3+
2bn = 3, but then it incorrectly factors the left side as n(an2+b). Also, since n and an2+b are both integers, n >0, and 3 is a prime number, it does follow (as
in the proof) that one of n, an2+b has to equal 3 and the other has to equal 1.
This proof only considers the case wheren = 3 andan2+b= 1. The n= 1 needs to be considered, and doing so is analogous to the n= 3 case.
The proof should proceed as follows. Rewrite an3+ 2bn= 3 as n(an2+ 2b) = 3.
Sincen and an2+ 2b are both integers whose product is 3, and since n is positive an2 + 2b must be a positive. Since 3 is prime, there are only two possibilities:
n= 3 or n= 1. If n= 3, then we have
and dividing both sides by 3, we obtain
9a+ 2b = 1. Ifn = 1, then we have
a+ 2b= 3. This completes the proof.
(b) This proof is incorrect, since it assumes the negation of the hypothesis of the state-ment. A correct proof by contradiction assumes the negation of the conclusion. See (a) for a correct proof.
§5.1: 1. (a) A=B. It is evident thatA⊆B. Ifx∈B, thenx= 3, x=−3, x= 2, orx=−2. So, B ⊆A.
(b) Yes.
(c) No, the set C is not equal to the set D. The set C is empty. The set D contains many, many real numbers.
(d) Yes; the empty set is a subset of any set.
(e) No, A contains negative numbers, e.g. −3, but D doesn’t.
3. The following statements are true. A ⊆ B, A ⊂ B, A 6=B. 5 ∈ C. A ⊆ C, A ⊂ C, A 6= C. {1,2} 6⊆ A, {1,2} 6= A. 4 6∈ B. card(A) = card(D). A ∈ P(A). ∅ ⊆ A, ∅ ⊂ A, ∅ 6= A. {5} ⊆ C, {5} ⊂ C, {5} 6= C. {1,2} ⊆ B, {1,2} ⊂ B, {1,2} 6= B. {3,2,1} ⊆ D, {3,2,1} ⊂ D, {3,2,1} 6= D. D 6⊆ ∅, D 6= ∅. card(A) < card(B), card(A)6= card(B). A∈P(B).
5. (a) It is not the case that {a, b} ⊆ {a, c, d, e}. This is because there is an element b∈ {a, b} which is not an element of {a, c, d, e}.
(b) This is true. The set {x ∈ Z|x2 < 5} is the set {−2,−1,0,1,2}. The even
members of this set are {−2,0,2}.
(c) This is true. The empty set is a subset of any set.
(d) This is not true. The set{a} is a subset of{a, b}, and so{a} is an element of the power set ofA.
6. (a) x6∈A∩B if and only ifx6∈A or x6∈B. (b) x6∈A∪B if and only ifx6∈A and x6∈B.
(c) x6∈A−B if and only ifx∈Ac orx∈B. 8. (a) A∩B ={7,9,11,13, . . .}.
(b) A∪B ={1,3,5,7,8,9,10,11, . . .}. (c) (A∪B)c={2,4,6}.
(d) Ac∩Bc={2,4,6}.
(f) (A∩C)∪(B∩C) ={3,9,12,15,18,21, . . .}. (g) B∩D={ }.
(h) (B ∩D)c ={1,2,3, . . .}. (i) A−D={7,9,11,13,15, . . .}. (j) B−D={1,3,5,7,9, . . .}.
(k) (A−D)∪(B−D) = {1,3,5,7,9, . . .}. (l) (A∪B)−D={1,3,5,7,9, . . .}.
9. (a) For all x∈U, if x∈P −Q then x∈R∩S.
(b) There is an x such that x∈P −Q and x6∈R∩S. (c) For all x∈U, if x6∈R∩S, then x6∈P −Q. 11. Picture rectangles with lots of circles inside them!
